### Vector calculus in plane and space: Bases, Coordinates and Equations

### Planes in coordinate space

We have seen previously that a parametric representation of a plane #V# in the space has the shape # \vec{a}+\lambda \cdot\vec{u}+ \mu\cdot \vec{v}#, wherein #\vec{a}#, #\vec{u}#, and #\vec{v}# are vectors, so that #\vec{u}# and #\vec{v}# are not on the same line as #\vec{0}#, and #\lambda# and #\mu# are parameters. The vector that is formed, represents a general element of the plane #V#. This representation can be written in coordinates #x_1#, #x_2#, and #x_3# in many ways:

- Parametric representation in 'row format'

\[

\rv{x_1, x_2, x_3} = \rv{a_1, a_2 , a_3}+ \lambda\cdot\rv{u_1, u_2, u_3} + \mu\cdot\rv{v_1, v_2, v_3}\tiny.

\] - Parametric representation in column format:

\[

\left(\begin{array}{l}

x_1 \\x_2 \\x_3

\end{array}\right)

=

\left(\begin{array}{l}

a_1 \\a_2 \\a_3

\end{array}\right)

+

\lambda \cdot

\left(\begin{array}{l}

u_1 \\ u_2 \\ u_3

\end{array}\right)

+

\mu \cdot

\left(\begin{array}{l}

v_1 \\v_2 \\v_3

\end{array}\right) \,\tiny.

\] - Or just as each coordinate separately:

\[

\begin{array}{lll}

x_1 & = & a_1 + \lambda \cdot u_1 + \mu\cdot v_1 \\

x_2 & = & a_2 + \lambda\cdot u_2 + \mu\cdot v_2 \\

x_3 & = & a_3 + \lambda\cdot u_3 + \mu\cdot v_3\tiny.\\

\end{array}

\]

From parametric representation of a plane to equation

Each parametric representation with parameters #\lambda# and #\mu# of a plane in the coordinate space #\mathbb{R}^3# can be seen as three linear equations in the parameters #\lambda# and #\mu# and the coordinates #x_1#, #x_2#, and #x_3#.

Eliminating the parameters #\lambda# and #\mu# from the three equations, results in a comparison of the plane, which is a linear equation in the coordinates #x_1#, #x_2#, and #x_3#: \[d_1\cdot x_1 + d_2\cdot x_2 + d_3\cdot x_3 = d_4\tiny,\] for definite #d_1#, #d_2#, #d_3#, and #d_4#. In this case, at least one of the coefficients #d_1#, #d_2#, #d_3# must be not equal to zero.

Below are examples of different methods to go from a parametric equation to an equation for a plane and back.

We are looking for an equation of the form #a\cdot x+b\cdot y+c\cdot z+d=0#, such that each point of the form \[\rv{x,y,z} = \left[ 1 , 0 , -1 \right] +\lambda\cdot \left[ -2 , -2 , 1 \right]+\mu\cdot \left[ -3 , 1 , 3 \right]\]is a solution. This means that #x#, #y#, and #z# satisfy the equations

\[\eqs{x&=&-2\cdot \lambda-3\cdot \mu+1\cr y&=&\mu-2\cdot \lambda\cr z&=&\lambda+3\cdot \mu-1\cr}\]

By means of

*elimination*we work towards a linear equation with unknown #x#, #y#, and #z# in which #\lambda# and #\mu# no longer occur. First we eliminate #\lambda# from these equations by taking suited linear combinations:

\[\begin{array}{rcl}\eqs{-2\cdot x+2\cdot y-8\cdot \mu+2&=&0\cr x+2\cdot z-3\cdot \mu+1&=&0\cr y+2\cdot z-7\cdot \mu+2&=&0\cr}

\end{array}\]Here we obtained the first equation by multiplying the first equation of the system above with #-2# (the coefficient of #\lambda# in the second equation), the second one with #-2# (the coefficient of #\lambda# in the first equation), subtrachting the second result from the first result, and moving all terms to the left hand side in the resulting equation. The second and third equation can be obtained in the same way, but then assuming two different equation from the three equations above with #\lambda# and #\mu#.

Next, we eliminate the parameter #\mu# with the same method. This gives

\[\begin{array}{rcl}\eqs{14\cdot x-6\cdot y+16\cdot z+2&=&0\cr -7\cdot x+3\cdot y-8\cdot z-1&=&0\cr 14\cdot x-6\cdot y+16\cdot z+2&=&0\cr}

\end{array}\]All three equations are equal to \[{7\cdot x-3\cdot y+8\cdot z +1=0}\], apart from a multiple. Hence, this is an equation of the plane.

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